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\begin{document}

\begin{center}
{\LARGE \bf SSF.Util.Random Design Notes}
\end{center}
\vspace{0.25cm}

\section{Transformation method}

If $p(x)$ is a continuous probability density function,
$$
F(y) = \int_{- \infty}^{y} p(s)ds
$$
and $x$ is uniformly distributed on $[0,1]$, then the transformed variable $y$
$$
y(x) = F^{-1}(x)
$$
has probability density $p(y)$.

\subsubsection{Exponential distribution}
\begin{eqnarray*}
p(y) & = & \frac{1}{\mu}e^{-y/\mu}, \; \; \mu > 0, \; \; y > 0 \\
E[y] & = & \mu \\
V[y] & = & \mu^2
\end{eqnarray*}

Then:
\begin{eqnarray*}
F(y) & = & 1 - e^{-y/\mu} \\
F^{-1}(x) & = & -\mu \ln(1 - x)
\end{eqnarray*}
and if $x$ is uniformly distributed on  $(0,1)$, so is $1-x$, and
thus
$$
y =  -\mu \ln(x)
$$
is exponentially distributed with mean $\mu$.

\subsubsection{Pareto distribution}

\begin{eqnarray*}
p(y) & = &
           \left\{
           \begin{array}{cl}
           \alpha k^{\alpha} y^{-(1+\alpha)}, & \mbox{$y \geq k > 0$} \\
           0, & \mbox{$y < k$}
           \end{array}
           \right. \\
E[y] & = & \frac{\alpha k}{\alpha -1}, \;\; \mbox{$\alpha > 1$} \\
V[y] & = & \frac{\alpha k^2}{(\alpha - 2)(\alpha -1)^2}, \;\; \mbox{$\alpha > 2$}
\end{eqnarray*}

Then
\begin{eqnarray*}
F(y) & = & 1 - (k/y)^{\alpha}, \\
F^{-1}(x) & = & k (1 -x)^{-1/\alpha}
\end{eqnarray*}

and if $x$ is uniformly distributed on  $(0,1)$, so is $1-x$, and
thus
$$
y =  k x^{-1/\alpha}
$$
is Pareto-distributed with {\em scale} $k$ and {\em shape} $\alpha$.


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